两道简单却实用的python面试题

List=['b','b','d','b','c','a','a'] 

print "the list is:" ,  List 

if List: 

        List.sort() 

        last = List[-1] 

        for i in range(len(List)-2, -1, -1): 

                if last==List[i]: 

                        del List[i] 

                else: 

                        last=List[i] 

print "after deleting the repeated element the list is : " , List

l1 = ['b','c','d','b','c','a','a'] 

l2 = [] 

[l2.append(i) for i in l1 if not i in l2] 

print l2 

题目三：实现斐波那契（Fibonacci）数列

方法一：递归

def fibonacci2(n): 

    if n == 1 or n == 2: 

        return 1 

    else: 

        return fibonacci2(n-1) + fibonacci2(n-2)

def fibonacci(n): 

    if n == 1 or n == 2: 

        return 1 

    nPre = 1 

    nLast = 1 

    nResult = 0 

    i = 2 

    while i < n: 

        nResult = nPre + nLast 

        nPre = nLast 

        nLast = nResult 

        i += 1 

    return nResult 

print fibonacci(5)